Wednesday, May 4, 2016

`3/(x^2 - 3x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`3/(x^2-3x)`


To decompose a fraction, factor the denominator.


`3/(x(x-3))`


Then,  write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction. 


`A/x`  and   `B/(x-3)`


Add these two fractions and set it equal to the given rational expression.


`3/(x(x-3)) = A/x + B/(x-3)`


To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.


`x(x-3)*3/(x(x-3))=(A/x+B/(x-3))*x(x-3)`


`3=A(x-3) +...

`3/(x^2-3x)`


To decompose a fraction, factor the denominator.


`3/(x(x-3))`


Then,  write a fraction for each factor. Since the numerator is still unknown, assign a variable to the numerator of each fraction. 


`A/x`  and   `B/(x-3)`


Add these two fractions and set it equal to the given rational expression.


`3/(x(x-3)) = A/x + B/(x-3)`


To get the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.


`x(x-3)*3/(x(x-3))=(A/x+B/(x-3))*x(x-3)`


`3=A(x-3) + Bx`


Then, plug-in the roots of each factor.


For the factor (x-3), its root is x=3.


`3=A(3-3) + B(3)`


`3=3B`


`3/3=(3B)/3`


`1=B`


For the factor x, its root is x=0.


`3=A(0-3)+B(0)`


`3=-3A`


`3/(-3)=(-3A)/(-3)`


`-1=A`


So the given rational expression decomposes to:


`-1/x + 1/(x-3)`


This can be re-written as:


`1/(x-3) - 1/x`



To check, express the two fractions with same denominators.


`1/(x-3)-1/x=1/(x-3)*x/x - 1/x*(x-3)/(x-3)=x/(x(x-3)) - (x-3)/(x(x-3))`


Now that they have same denominators, proceed to subtract them.


`=(x-(x-3))/(x(x-3)) = (x - x + 3)/(x(x-3))=3/(x(x-3))=3/(x^2-3x)`



Therefore,  `3/(x^2-3x) = 1/(x-3)-1/x` .

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Exposition A "decidedly amused" Bobby Gillian leaves the offices of Tolman & Sharp where he is given an envelope containing $1...